a ball is thrown downward from the top of a buliding with an initial speed of 25 m/s. it strikes the ground after 2.0 seconds. how high is the building.
A) 50 m
B)70 m
C)30 m
D) 20 m
2007-01-26 05:58:03 · 3 answers · asked by Julie Kim 1 in Science & Mathematics ➔ Engineering
The equation to use for this problem is:
y=vo*(t)+1/2 *a*(t^2)
where y is the displacement, vo is the initial velocity, t is the time, and a is the acceleration (in this case, gravity). Since this is all only 1 direction, its a fairly simple problem, just plug in the appropriate numbers! (Note that if the ball is thrown upward or if acceleration is working in the opposite direction, you would need to include negative signs. For more advanced problems, a coordinate system is a must)
vo = 25 m/s
a = gravity = ~9.81m/(s^2)
t = 2s
y=25*2+1/2 *9.81*(2^2)
y = 69.62 = ~70m
answer B -> 70m
2007-01-26 06:54:45 · answer #1 · answered by Nes 3 · 0⤊ 0⤋
B.
x = Vt + 1/2at^2
x = 25*2 + 1/2*10*4
x = 70
2007-01-26 06:04:50 · answer #2 · answered by gebobs 6 · 0⤊ 0⤋
x=vo*(t)+1/2 *a* t^2
2007-01-26 06:09:45 · answer #3 · answered by BRUZER 4 · 0⤊ 0⤋